[LeetCode] 507. Perfect Number 完美数字
We define the Perfect Number is a positive integer that is equal to the sum of all its positive divisors except itself.
Now, given an integer n, write a function that returns true when it is a perfect number and false when it is not.
Example:
Input: 28 Output: True Explanation: 28 = 1 + 2 + 4 + 7 + 14
Note: The input number n will not exceed 100,000,000. (1e8)
定义Perfect Number是一个正整数，它等于除了它自己之外的所有正除数的总和。现在，给定一个整数n，编写一个函数，当它是一个完美数字时返回true，否则返回false。
解法：直接解就可以。注意处理num是1的时候。还有技巧是，只计算num/2就可以，如果能整除就把除数和结果都累加。
Java:
public class Solution {
public boolean checkPerfectNumber(int num) {
if (num == 1) return false;
int sum = 0;
for (int i = 2; i <= Math.sqrt(num); i++) {
if (num % i == 0) {
sum += i + num / i;
}
}
sum++;
return sum == num;
}
}
Python:
class Solution(object):
def checkPerfectNumber(self, num):
"""
:type num: int
:rtype: bool
"""
if num <= 0: return False
ans, SQRT = 0, int(num ** 0.5)
ans = sum(i + num//i for i in range(1, SQRT+1) if not num % i)
if num == SQRT ** 2: ans -= SQRT
return ans - num == num
Python:
class Solution(object):
def checkPerfectNumber(self, num):
"""
:type num: int
:rtype: bool
"""
if num == 1:
return False
i = 2
s = 1
while i * i < num:
if num % i == 0:
s += i
s += num / i
i += 1
return True if s == num else False
C++:
class Solution {
public:
bool checkPerfectNumber(int num) {
if (num == 1) return false;
int sum = 1;
for (int i = 2; i * i <= num; ++i) {
if (num % i == 0) sum += (i + num / i);
if (i * i == num) sum -= i;
if (sum > num) return false;
}
return sum == num;
}
};
C++:
class Solution {
public:
bool checkPerfectNumber(int num) {
return num==6 || num==28 || num==496 || num==8128 || num==33550336;
}
};
